76.4. THE GENERAL CASE 2559
Now the right side converges to 0 as r,q→ ∞ and so there is a subsequence, denoted withthe index k such that if p > k,
E
(sup
t∈[0,T ]
⟨Buk−Bup,uk−up
⟩(t)
)≤ 1
2k (76.4.34)
Then consider the earlier local martingales. One of these is of the form
Mk =∫ t
0
(Φk ◦ J−1)∗Buk ◦ JdW
Then by the Burkholder Davis Gundy inequality and modifying constants as appropriate,
E((Mk−Mk+1)
∗)≤ C
∫Ω
(∫ T
0
∥∥∥(Φk ◦ J−1)∗Buk−(Φk+1 ◦ J−1)∗Buk+1
∥∥∥2dt)1/2
dP
≤C∫
Ω
( ∫ T0 ∥Φk−Φk+1∥2 ⟨Buk,uk⟩
+∥Φk+1∥2 ⟨Buk−Buk+1,uk−uk+1⟩dt
)1/2
dP
≤ C∫
Ω
(∫ T
0∥Φk−Φk+1∥2 ⟨Buk,uk⟩dt
)1/2
+C∫
Ω
(∫ T
0∥Φk+1∥2 ⟨Buk−Buk+1,uk−uk+1⟩dt
)1/2
dP
≤C∫
Ω
supt⟨Buk,uk⟩1/2
(∫ T
0∥Φk−Φk+1∥2 dt
)1/2
dP
+C∫
Ω
supt⟨Buk−Buk+1,uk−uk+1⟩1/2
(∫ T
0∥Φk+1∥2 dt
)1/2
dP
≤C(∫
Ω
supt⟨Buk,uk⟩dP
)1/2(∫Ω
∫ T
0∥Φk−Φk+1∥2 dtdP
)1/2
+C(∫
Ω
supt⟨Buk−Buk+1,uk−uk+1⟩dP
)1/2(∫Ω
∫ T
0∥Φk+1∥2 dtdP
)1/2
From the above inequalities, after adjusting the constants, the above is no larger than anexpression of the form C
( 12
)k/2which is a summable sequence. Then
∑k
∫Ω
supt∈[0,T ]
|Mk (t)−Mk+1 (t)|dP < ∞
Then {Mk} is a Cauchy sequence in M1T and so there is a continuous martingale M such
that
limk→∞
E(
supt|Mk (t)−M (t)|
)= 0 (76.4.35)